joshgarrod Posted January 12, 2011 Share Posted January 12, 2011 Hi all, I have an issue with an image upload script. So far the script will upload the image to the server and rename it with a random 5 digit number on the end. I then INSERT this data into a table for sourcing the image later on. the only problem is that the data in the table is the image's original name not the new one. how do I get it to change that name too? Thanks. <?php $idir = "uploads/"; // Path To Images Directory if (isset ($_FILES['fupload'])){ //upload the image to tmp directory $url = $_FILES['fupload']['name']; // Set $url To Equal The Filename For Later Use if ($_FILES['fupload']['type'] == "image/jpg" || $_FILES['fupload']['type'] == "image/jpeg" || $_FILES['fupload']['type'] == "image/pjpeg") { $file_ext = strrchr($_FILES['fupload']['name'], '.'); // Get The File Extention In The Format Of , For Instance, .jpg, .gif or .php $copy = copy($_FILES['fupload']['tmp_name'], $idir. basename($_FILES['fupload']['name'], $file_ext).rand(10000 , 99999).$file_ext); // Move Image From Temporary Location To Perm } } error_reporting (E_ALL ^ E_NOTICE); $usr = "user"; $pwd = "pword"; $db = "db"; $host = "host"; # connect to database $cid = mysql_connect($host,$usr,$pwd); if (!$cid) { echo("ERROR: " . mysql_error() . "\n"); } if ($_POST['submit']) { $logo = mysql_real_escape_string("$idir" . $_FILES['fupload']['name']); $SQL = " INSERT INTO mhhire "; $SQL .= " (logo) VALUES "; $SQL .= " ('$logo') "; $result = mysql_db_query($db,$SQL,$cid); $last=mysql_insert_id(); if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n"); } header("location:thanks.php"); exit(); } ?> Quote Link to comment Share on other sites More sharing options...
litebearer Posted January 12, 2011 Share Posted January 12, 2011 set the $logo value just BEFORE you 'copy' the temp image. then use the name for both the destination copy as well as the db data Make sense? Quote Link to comment Share on other sites More sharing options...
joshgarrod Posted January 12, 2011 Author Share Posted January 12, 2011 Hi, thanks for the reply... Im sorry I do not quite understand how you mean? Quote Link to comment Share on other sites More sharing options...
litebearer Posted January 12, 2011 Share Posted January 12, 2011 Take a look here, it should make it clearer... http://www.phpeasystep.com/phptu/18.html <?php // Your file name you are uploading $file_name = $HTTP_POST_FILES['ufile']['name']; // random 4 digit to add to our file name // some people use date and time in stead of random digit $random_digit=rand(0000,9999); //combine random digit to you file name to create new file name //use dot (.) to combile these two variables $new_file_name=$random_digit.$file_name; //set where you want to store files //in this example we keep file in folder upload //$new_file_name = new upload file name //for example upload file name cartoon.gif . $path will be upload/cartoon.gif $path= "upload/".$new_file_name; if($ufile !=none) { if(copy($HTTP_POST_FILES['ufile']['tmp_name'], $path)) { echo "Successful<BR/>"; //$new_file_name = new file name //$HTTP_POST_FILES['ufile']['size'] = file size //$HTTP_POST_FILES['ufile']['type'] = type of file echo "File Name :".$new_file_name."<BR/>"; echo "File Size :".$HTTP_POST_FILES['ufile']['size']."<BR/>"; echo "File Type :".$HTTP_POST_FILES['ufile']['type']."<BR/>"; } else { echo "Error"; } } ?> Quote Link to comment Share on other sites More sharing options...
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